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138 lines
2.5 KiB
138 lines
2.5 KiB
#include "stdafx.h"
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#include "BranchPoint.h"
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#include <cmath>
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#include <vector>
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#include <queue>
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#include <map>
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#include <iostream>
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#include <fstream>
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#include <sstream>
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#include <algorithm>
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using namespace std;
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// 简单的路径信息,路径规划的直接结果
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struct Path
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{
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vector<string> wirelist; // 路径中的线缆名
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vector<P> points; // 路径上的点,包括卡箍和分支点
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vector<int> inOut; // 卡箍方向是正向为0,反向为1,分支点为-1
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double dia; // 路径直径
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Path()
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{
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dia = 0;
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}
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// 返回path中点的数量
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inline int size()
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{
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assert(inOut.size() == points.size());
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return points.size();
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}
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// 把路径中经过的卡箍和分支点都标记成使用过的状态
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void markPoint()
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{
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for (int i = 0; i < points.size(); i++)
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{
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P &tp = points[i];
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if (tp.type == 1 || tp.type == 2)
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{
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BranchPoint *pb = branchPointSet.getBranchPointPointer(tp.ref);
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pb->used = true;
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pb->pc1->used = true;
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pb->pc2->used = true;
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}
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else if (tp.type == 0)
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{
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clipSet.c[tp.ref].used = true;
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}
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}
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}
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// 固定所有经过的分支点
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void fixBranchPoint()
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{
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for (int i = 0; i < points.size(); i++)
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{
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P &tp = points[i];
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if (tp.type == 1)
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{
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BranchPoint *pb = branchPointSet.getBranchPointPointer(tp.ref);
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pb->fix();
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}
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}
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}
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// 检查路径合法性
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void illegalCheck()
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{
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map<P, int> vis;
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for (int i = 0; i < points.size(); i++)
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{
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assert(vis[points[i]] == 0);
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vis[points[i]] = 1;
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if (i == 0 || i == points.size() - 1)
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{
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assert(points[i].type == 3);
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}
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if (points[i].type != 0)
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{
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assert(inOut[i] < 0);
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}
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}
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}
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// 将路径中的点翻转成正常反向,并生成inOut数组
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void processDir()
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{
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for (int j = 0; j < points.size(); j++)
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{
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if (points[j].type != 0)
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{
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inOut.push_back(-1);
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continue;
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}
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double reverse = 0;
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P dir = dirP(points[j]);
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if (j != 0)
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{
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reverse += pi / 2 - Angle(points[j] - points[j - 1], dir);
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}
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if (j != points.size() - 1)
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{
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reverse += pi / 2 - Angle(points[j + 1] - points[j], dir);
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}
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if (reverse < 0)
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{
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// points[j].reverse();
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inOut.push_back(1);
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}
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else
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inOut.push_back(0);
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}
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}
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// 在路径确定后,处理一些能在pass层面处理的信息
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void postProcess()
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{
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this->fixBranchPoint();
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// this->processDir();
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this->markPoint();
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}
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// 翻转整条路径
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void reverse()
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{
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std::reverse(points.begin(), points.end());
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assert(inOut.size() == points.size());
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for (int i = 0; i < points.size(); i++)
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{
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if (inOut[i] != -1)
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{
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inOut[i] = inOut[i] ^ 1;
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points[i].reverse();
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}
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}
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}
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};
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